# normal subgroup problem

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Let G be a group in which, (ab)^2 = (a^2)(b^2) for

all a,b ∈G . Show that H = { g^2|g ∈G } is
a normal subgroup of G.

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Where are you stuck, showing that H is a subgroup or showing that H is normal?

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Actually both. I can prove it by using axioms & definitions, 1st by showing that H is a sgp & then its normal. But the question holds only 2 marks. So I think there is a shorter way of doing it as well. I want help regarding that. Thank you.

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I suspect you are supposed to see that G is abelian. That certainly helps for showing normality.

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Okay. Thank you Sir.

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Posted (edited)

Deleted

Edited by Martin Rattigan
was rubbish

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Posted (edited)
On 12/20/2018 at 9:10 AM, taeto said:

I suspect you are supposed to see that G is abelian. That certainly helps for showing normality.

Good tip.+1. Is this homework, @Prasant36?

Edit: You also need Abelian character for showing closure

Edited by joigus

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Posted (edited)
16 hours ago, joigus said:

Edit: You also need Abelian character for showing closure

in mathematics, closure can correspond many things.may I ask:  which type of closure do you meantion here?

Edited by ahmet

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9 minutes ago, ahmet said:

in mathematics, closure can correspond many things.may I ask:  which type of closure do you meantion here?

Take a guess.

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21 minutes ago, joigus said:

Take a guess.

I can guess many things really such as Algebraic closure, closure in topology and analysis , and functional analysis...

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1 minute ago, ahmet said:

Algebraic closure,

Bingo!!

"Is a group" refers to algebraic properties.

$g_{1},g_{2}\in H\Rightarrow g_{1}g_{2}\in H$

We're not talking topological groups. (I'm not aware that anybody mentioned a basis of neighbourhoods). Welcome to page 1.

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I think almost all parts of mathematics have intersections (even topology and functional analysis with algebra)

....

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1 hour ago, ahmet said:

in mathematics, closure can correspond many things.may I ask:  which type of closure do you meantion here?

Also, by "normal" (in this context) I understand:

$gHg^{-1}\subseteq H$

Not "perpendicular". Any more questions?

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Just now, joigus said:

Also, by "normal" (in this context) I understand:

gHg1H

Not "perpendicular". Any more questions?

no (more) questions ,I just tried to understand what you meant

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Just now, ahmet said:

I think almost all parts of mathematics have intersections (even topology and functional analysis with algebra)

....

They do. I know, and you know. And I know you know. And you know I know you know.

Can we stick to the topic, please?   It's algebra. Group theory. That's why we are @

1 minute ago, ahmet said:

no (more) questions ,I just tried to understand what you meant

Ah, OK. I'm sorry if I misunderstood your question in any sense.

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